- #1
pivoxa15
- 2,255
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How would I go about finding a number that is a primitive root modulo 125?
There definitely exists a primitive root since 5^3 =125
The problem basically comes down to finding 'a' (primitive root);
a^50 congruent to -1 (mod 125)
Anyone know a way apart from trial and error?
Thanks
There definitely exists a primitive root since 5^3 =125
The problem basically comes down to finding 'a' (primitive root);
a^50 congruent to -1 (mod 125)
Anyone know a way apart from trial and error?
Thanks